CF#333（Div2）B. Approximating a Constant Range（RMQ）

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B. Approximating a Constant Range

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data
points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an.
There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if
the difference between the largest and the smallest value in that range is at most 1. Formally, let M be
the maximum and m the minimum value of ai for l ≤ i ≤ r;
the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) —
the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Examples

input

5
1 2 3 3 2

output

4

input

11
5 4 5 5 6 7 8 8 8 7 6

output

5

Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10];
the only almost constant range of the maximum length 5 is [6, 10].

题目大意：求序列最大的数与最小的数相差不超过1的区间最大长度。

解题思路：RMQ。

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┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;

#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")

int a[100005];
int mx[100005][20],mi[100005][20];

void RMQ(int n)
{
for(int i=1;i<=n;i++)
mx[i][0] = mi[i][0] = a[i];
for(int j=1;(1<<j)<=n;j++)
{
for(int i=1;i+(1<<j)-1<=n;i++)
{
mx[i][j] = max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);
mi[i][j] = min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
}
}
}

int query(int l,int r)
{
int k = 0;
while((1<<(k+1))<(r-l+1))k++;
int ans1 = max(mx[l][k],mx[r-(1<<k)+1][k]);
int ans2 = min(mi[l][k],mi[r-(1<<k)+1][k]);
return ans1-ans2;
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
int n;
n = read();
for(int i=1;i<=n;i++)a[i] = read();
RMQ(n);
int p = 1;
int ans = 0;
for(int i=1;i<=n;i++)
{
while(p<=i && query(p,i)>1)
{
p++;
}
ans = max(ans,i-p+1);
}
printf("%d\n",ans);

return 0;
}